Transcript Chapter 10 Linear Programming

Linear Programming
Linear programming
A technique that allows decision
makers to solve maximization and
minimization problems where there are
certain constraints that limit what can
be done, given that all objectives and
constraints are linear.
Definitions
 The
objective function is the function to
be maximized (or minimized)
 The
constraints are given in inequalities.
 Both
the objective and constraints are
linear in choice variables
Example

Process 1: requires 3 machine-hours and 0.4
labor-hours per batch of cloth.

Process 2: requires 2.5 machine-hours and 0.5
labor-hours per batch of cloth.

Process 3: requires 5.25 machine-hours and
0.35 labor-hours per batch of cloth.

The profit is $1, $0.9, and $1.1 with process 1, 2,
and 3, respectively.

Max L and K are 600 and 6,000, respectively.
The objective is to maximize profit
subject to resource constraints
 Maximize P = 1.0Q1 + 0.9Q2 + 1.1Q3
subject to the constraints:
3Q1 + 2.5Q2 + 5.25Q3 ≦ 6000 (Machine hours)
0.4Q1 + 0.5Q2 + 0.35Q3 ≦ 600 (Labor Hours)
Q1 ≧ 0; Q2 ≧ 0; Q3 ≧ 0.
Machine Hours
Process rays
15000
Q3 (5.25 / 0.35)
Q1
10000
Q1 (3 / 0.4)
5000
Q2 (2.5 / 0.5)
0
0
500
Labor Hours
1000
Q2
Q3
Machine Hours
Isoquants (line ABC, NOT line AC)
15000
Q3 (5.25 / 0.35)
10000
A
5000
Q1 (3 / 0.4)
B
C
0
0
500
Labor Hours
Q2 (2.5 / 0.5)
1000
Q1
Q2
Q3
Machine Hours
Feasible Set
15000
Q3 (5.25 / 0.35)
Q1
10000
(6000, 600)
5000
Q1 (3 / 0.4)
Q2
Q3
Q2 (2.5 / 0.5)
0
0
500
Labor Hours
1000
Machine Hours
Graphical Solution: A*
15000
Q3 (5.25 / 0.35)
Q1
10000
A*
Q1 (3 / 0.4)
5000
Q2
Q3
Q2 (2.5 / 0.5)
0
0
500
Labor Hours
1000
 Suppose
that the company is no longer
constrained by limits on the amount of Labor
and capital.

It can hire all of labor it wants at $12 per hour and
all of the capital at $1.6 per machine-hour.

Its problem is to choose that combination of
processes which will produce, say, 400 units at
minimum cost.

Min
C = 9.6 Q1 + 10 Q2 + 12.6 Q3
Subject to:
Q1 + Q2 + Q3 ≧ 400
Q1 ≧ 0, Q2 ≧ 0, Q3 ≧ 0
Machine Hours
Optimal Solution:
Cost Minimization Problem
15000
Iso-cost line
Q3 (5.25 / 0.35)
Q1
10000
Q1 (3 / 0.4)
5000
Q2 (2.5 / 0.5)
0
0
500
B*
Q2
Q3
1000
Labor Hours
Solution: All of the output are produced with process 1
Simplex method
A systematic process for comparing
extreme point or corner solutions to
linear programming problems
Dual problems
 Every
optimization problem has a
corresponding problem called its dual
 If
the primal problem is a maximization,
the dual is a minimization (and vice versa)
 Solutions
to the dual are shadow prices
Shadow Prices
 Shadow
prices tell you what would happen to
the objective function if you relaxed the
constraint by one unit.
 They
show which types of capacity are
bottlenecks, or effective constraints on output,
since capacity that is underutilized receives a
zero shadow price.
 More
important, shadow prices indicate how
much it would be worth to management to
expand each type of capacity.
Example
 Output 1 (Q1 ): requires 3 units of capital and 2
units of labor to produce one unit of output 1.
 Output 2 (Q2 ): requires 5 units of capital and 4
units of labor to produce one unit of output 2.
 The firm has 4,000 units of labor and 5,400 units
of capital.
 The price of Q1 and Q2 are $50 and $80,
respectively.
Primal problem
 The
primal problem is to find the value of Q1
and Q2 that maximize total revenue subject
to the resource constraint.

Max
Subject to:
TR = 50 Q1 + 80 Q2
2Q1 + 4 Q2 ≦ 4,000
3Q1 + 5 Q2 ≦ 5,400
Q1 ≧ 0, Q2 ≧ 0
Dual problem
The dual problem is to find the value of PL
and PK , which are the price of L and K,
respectively, to minimize total value of
resource available subject to the constraint.

Min
C = 4,000 PL + 5,400 PK
Subject to:
2 PL + 3 PK ≧ 50
(*)
4 PL + 5 PK ≧ 80
(**)
PL ≧ 0, PK ≧ 0
 Constraint (*) and (**) say that the total value of
resources used in the production of a unit of
output must be greater or equal to the price.
PL and PK are the prices that a manager should
be willing to pay for these resources.
 They are the opportunity cost of using theses
resources.
 If a resource is not fully utilized, its shadow prices
will be zero, since an extra unit of the resource
would not increase total revenue (objective
function).
Max
S. t.
50 Q1 +
80
Q2
2 Q1 +
4
Q2 ≦ 4000
3 Q1 +
5
Q2 ≦ 5400 PK
PL
Slack Variables
 Slack variables (UL and UK ) represent the amounts
of various inputs that are unused.

2Q1 + 4 Q2 + UL
3Q1 + 5 Q2 + UK
=
=
4,000
5,400
 If the slack variable for an input turns out to be zero,
this means that this input is fully utilized
(Uh = 0 implies Ph > 0 , h = L, K )
 If it turns out to be positive, this means that some of
this input is redundant
(Uh > 0 implies Ph = 0 , h = L, K ) .
How H. J. Heinz Minimize
Its Shipping

To determine how much ketchup each factory
should send to each warehouse, Heinz has used
linear programming techniques.

The capacities of each factory, the requirements
of each warehouse, and freight rates are given in
the first table.

The optimal daily shipment from each factory to
each warehouse is shown in the second table.
For example, all warehouse A’s ketchup should
come from factory I.
Freight rates (cents per cwt.) from factory:
Warehouse
I
II
III
IV
A
16
16
6
13
B
20
18
8
10
C
30
23
8
9
D
10
15
10
8
E
31
23
16
10
F
24
14
19
13
G
27
23
7
11
H
34
25
15
4
J
38
29
17
11
K
42
43
21
22
L
44
49
25
23
M
49
40
29
21
N
56
58
36
37
P
59
57
44
33
Q
68
54
40
38
R
66
71
47
43
S
72
58
50
51
T
74
54
57
55
U
71
75
57
60
Y
73
72
63
56
Daily
capacity 10,000 9,000
3,000
2,700
V
VI
24
22
14
10
10
13
23
27
16
16
18
10
6
5
8
16
20
26
30
37
500
VII
13
11
7
15
16
14
8
15
27
10
6
15
25
21
24
33
42
53
44
49
1,200
VIII
6
8
9
13
20
18
16
11
17
21
13
14
8
6
7
12
22
26
30
40
700
IX
31
29
22
19
14
9
6
9
19
18
19
21
19
10
19
26
16
19
30
31
300
X
37
33
29
19
17
14
10
16
8
24
15
12
9
8
10
19
15
14
41
31
500
XI
34
25
20
15
17
13
11
17
18
16
12
29
21
33
23
20
13
7
8
10
1,200
37
35
38
28
25
29
16
13
19
17
10
14
15
15
23
25
20
15
23
8
2,000
Daily
requirements
XII
40 (cwt.)
1,820
38
1,530
35
2,360
34
100
28
280
25
730
28
940
16
1,130
11
4,150
15
3,700
13
2,560
20
1,710
26
580
18
30
23
2,840
31
1,510
21
970
6
5,110
37
3,540
25
4,410
8,900
40,000
Factory
Warehouse
A
B
C
D
E
F
G
H
J
K
L
M
N
P
Q
R
S
T
U
Y
Total
I
1,820
1,530
II
III
IV
V
VI
VII
VIII
IX
X
XI
XII
2,360
100
280
730
940
1,130
4,150
700
1,360
3,000
1,200
140
1,570
580
30
1,340
500
500
810
500
700
90
2,160
10,000
880
5,110
180
9,000
3,000
2,700
500
1,200
700
300
1,200
500
1,200
2,000
2,000
2,410
8,900
Total
1,820
1,530
2,360
100
280
730
940
1,130
4,150
3,700
2,560
1,710
580
30
2,840
1,510
970
5,110
3,540
4,410
40,000